Question

Triangle ABC has vertices at A (1,-3), B (3,0)x and C (5,-3). What is the distance from vertex B to the midpoint of AC?

Answer 1

The distance from vertex B to the midpoint of AC is 3 units

Explanation:

we are given a triangle with vertices A (1,-3), B (3,0) and C (5,-3).

Mid point of of AC (x,y) = ( 1+5/2 , -3 -3/2)

= ( 6/2 , -6/2)

= (3 , -3)

Distance formula =
`√((x2 - x1)^2 + (y2 - y1)^2)`

=
`√((3-3)^2 + (0 + 3) ^2)`

=
`√((3)^2)`

= 3 units